Work-Rate Problems ③ Changing Workers Partway Through
Let us learn slightly more complicated patterns, such as “A works for N days and then B takes over” and “A and B take turns working each day.”
Work-Rate Problems with Alternating Days
Example Problem
A can finish a certain job alone in 6 days, and B can finish it alone in 4 days. A works on day 1, B on day 2, A on day 3, and so on, taking turns. On which day will the job be completed?
Solution:
① Set the total amount as 12, the least common multiple of 6 and 4.
② A’s one-day work = 12 ÷ 6 = 2, B’s one-day work = 12 ÷ 4 = 3
③ Amount completed in one “A day + B day” two-day set: 2 + 3 = 5
④ 12 ÷ 5 = 2 remainder 2 → In 2 sets (4 days), 5 × 2 = 10 units are completed. Remaining amount = 12 − 10 = 2
⑤ Day 5 is A’s turn (A completes 2 units in one day) → The remaining 2 units are completed exactly!
Answer: Day 5
A can finish a certain job alone in 6 days, and B can finish it alone in 4 days. A works on day 1, B on day 2, A on day 3, and so on, taking turns. On which day will the job be completed?
Solution:
① Set the total amount as 12, the least common multiple of 6 and 4.
② A’s one-day work = 12 ÷ 6 = 2, B’s one-day work = 12 ÷ 4 = 3
③ Amount completed in one “A day + B day” two-day set: 2 + 3 = 5
④ 12 ÷ 5 = 2 remainder 2 → In 2 sets (4 days), 5 × 2 = 10 units are completed. Remaining amount = 12 − 10 = 2
⑤ Day 5 is A’s turn (A completes 2 units in one day) → The remaining 2 units are completed exactly!
Answer: Day 5
The Key Is to Compare the “Remaining Amount” with the “One-Day Work”
After repeating the sets, if the remaining amount is less than the next person’s one-day work, the job will be completed partway through that day (the wording may differ depending on the problem).
Key Points
- First calculate the amount completed in one set, with A and B each working once.
- Repeat the sets until just before the total amount is completed, then handle the remainder one day at a time.
- When asked “On which day will it be completed?”, answer with the total number of days counted from the first day.
- If the job is completed partway through a day, the answer may be given as “○ days and ○ hours.”
Practice Problems
- A can finish a certain job alone in 10 days, and B can finish it alone in 15 days. If they work alternately—A on day 1, B on day 2, A on day 3, and so on—on which day will the job be completed?
- A certain job takes A 8 days working alone. After A works for N days, B takes over and finishes the remainder in 6 days. The job takes exactly 10 days in total. Find N.
- A and B work on a certain job alternately, one day each (A on day 1, B on day 2, and so on). A takes 12 days working alone, and B takes 6 days working alone. On which day will the job be completed?
Answers and Explanations
- Answer: Day 25
Explanation: Set the total amount as 30 (the LCM of 10 and 15). A’s one-day work = 3, and B’s one-day work = 2. The amount completed in 2 days (1 set) = 3 + 2 = 5. 30 ÷ 5 = 6 sets (12 days), which completes the entire job exactly. Counting the day: the job is completed on day 12. Check: A (odd-numbered days) works 6 times × 3 = 18, and B (even-numbered days) works 6 times × 2 = 12. 18 + 12 = 30✓. Answer: day 12. - Answer: N = 4
Explanation: Consider a job that takes A 8 days working alone. Set the total amount as 8, so A completes 1 unit per day. The amount A completes in N days = N. Remaining amount = 8 − N. B completes the remainder in 6 days: B’s one-day work = (8 − N) ÷ 6. Since the total time is exactly 10 days (N days of A’s work + 6 days of B’s work = 10), N + 6 = 10 → N = 4. Check: A works for 4 days and completes 4 units. Remaining amount = 8 − 4 = 4. B completes 4 units in 6 days → B’s one-day work = 4 ÷ 6 = 2/3. N = 4✓ - Answer: Day 8
Explanation: Set the total amount as 12 (the LCM of 12 and 6). A’s one-day work = 1, and B’s one-day work = 2. The amount completed in 2 days (1 set) = 1 + 2 = 3. 12 ÷ 3 = 4 sets (8 days), so the job is completed exactly. It is completed on day 8, B’s turn. Check: A (odd-numbered days: days 1, 3, 5, and 7) works 4 times × 1 = 4, and B (even-numbered days: days 2, 4, 6, and 8) works 4 times × 2 = 8. 4 + 8 = 12✓. Answer: day 8.